Question

if $\large\rm { \normal{f} \ : \ \mathbb{R}^{n} \longrightarrow \mathbb{R}^{m}}$ is differentiable at $\large\rm { p}$ , then $f$ is continuous at p.​

Answer 1

Correct Question:

if $\large { f \ : \ \mathbb{R}^{\rm{n}} \longrightarrow \mathbb{R}^{\rm{m}}}$ is differentiable at $\large\rm { p}$ , then prove that $f$ is continuous at p.

Answer:

from continuity of linear transformations,

$\rm{ \displaystyle\lim_{x \to p} f\rm{(x) }= \displaystyle\lim_{\rm{x \to p}} \rm{f(p) + Df(p)(x-p) + E(x-p)}}$

$\large\rm { = f(p) + Df(p)0 + \displaystyle\lim_{x \to p} \rm{E(x-p)}}$

Since f is differentiable at p, we can dismiss the last limit.

$\large\rm { \displaystyle\lim_{x \to p} \rm{ E(x-p) } = \displaystyle\lim_{x \to p} \rm{ || x - p || \frac{E(x-p)}{||x-p||}}}$

$\large\rm { = 0.0}$

use magnitude function $\large\rm { x \longrightarrow ||x||}$

now we conclude that

$\large\rm { \displaystyle\lim_{x \to p} f(x) = f(p)}$

$\large\rm { \therefore }$ F is continuous at p.